JEE Main 2024Inverse Trigonometric FunctionsInverse Trigonometric FunctionsMediumQuestion If y=cos(π3+cos−1x2), then (x−y)2+3y2 is equal to \text { If } y=\cos \left(\frac{\pi}{3}+\cos ^{-1} \frac{x}{2}\right) \text {, then }(x-y)^2+3 y^2 \text { is equal to } If y=cos(3π+cos−12x), then (x−y)2+3y2 is equal to Answer: 3Hide SolutionSolutiony=cos(π3+cos−1x2)=cos(π3)cos(cos−1(x2))−sin(π3.)sin(cos−1(x2))=12⋅x2−32⋅1−x24⇒4y=x−34−x2⇒(4y−x)2=3(4−x2)⇒16y2+x2−8xy=12−3x2x2+4y2−2xy=3(x−y)2+3y2=3\begin{aligned} & y=\cos \left(\frac{\pi}{3}+\cos ^{-1} \frac{x}{2}\right) \\ & =\cos \left(\frac{\pi}{3}\right) \cos \left(\cos ^{-1}\left(\frac{x}{2}\right)\right)-\sin \left(\frac{\pi}{3 .}\right) \sin \left(\cos ^{-1}\left(\frac{x}{2}\right)\right) \\ & =\frac{1}{2} \cdot \frac{x}{2}-\frac{\sqrt{3}}{2} \cdot \sqrt{1-\frac{x^2}{4}} \\ & \Rightarrow 4 y=x-\sqrt{3} \sqrt{4-x^2} \\ & \Rightarrow(4 y-x)^2=3\left(4-x^2\right) \\ & \Rightarrow 16 y^2+x^2-8 x y=12-3 x^2 \\ & x^2+4 y^2-2 x y=3 \\ & (x-y)^2+3 y^2=3 \end{aligned}y=cos(3π+cos−12x)=cos(3π)cos(cos−1(2x))−sin(3.π)sin(cos−1(2x))=21⋅2x−23⋅1−4x2⇒4y=x−34−x2⇒(4y−x)2=3(4−x2)⇒16y2+x2−8xy=12−3x2x2+4y2−2xy=3(x−y)2+3y2=3